November 8th 10th 2006

1
COE 328 Digital Systems and MicroprocessorsLab
7 Final ProjectProgrammable Processor Module
(PPM)

• Raymond Phan
• ENG 460
• http//www.ee.ryerson.ca/rphan

2
Outline of Lecture

• Objectives
• Introduction
• Levels of abstraction
• Basic Computer Architecture
• Overview of Customized Ryerson Microprocessor
  (CRM)
• Datapath Unit on CRM
• Architecture
• Control Signals

3
Outline of Lecture (2)

• Instruction set on CRM
• Details on each instruction available
• Control Unit on CRM
• Architecture
• Control Signals
• Procedure for Final Project
• Preparing EPROM Files
• Microinstructions ? EPROM Format
• Microinstruction Tables and Examples
• Format for code and addressing
• Conversion of EPROMs ? .asc to .hex

4
Outline of Lecture (3)

• What youre required to do
• Whats due for marks
• Format of Formal Lab Report
• End thank goodness right?

5
Objectives

• Get familiar with the basic elements that make up
  a microprocessor
• Working with a microprocessor that Ryerson Profs
  designed many years ago
• Derive the behaviour of the instruction set for a
  microprocessor (more on this later)

6
Objectives (2)

• Understand programs at the microprocessor level
• You get to code your very own custom program at
  this level!
• Things you need to know before we start
• Absolutely NO VHDL seen in this lab (yay?)
• No more work on the CPLDs for this lab (bigger
  yay?)
• Electrical guys ? No more VHDL for the rest of
  your life (YAY!) unless you want to get into
  digital design

7
Objectives (3)

• Also dont be scared! You might be scared of
  this project for the following reasons
• First time youll be programming at a lower level
  (lower than C or Java)
• First time youll see the things that make up any
  microprocessor that exists today
• You will be required to understand how exactly a
  microprocessor works.

8
Outline of Lecture

• Objectives
• Introduction
• Levels of abstraction
• Basic Computer Architecture
• Overview of Customized Ryerson Microprocessor
  (CRM)
• Datapath Unit on CRM
• Architecture
• Control Signals

9
Introduction Levels of Abstraction

• Microprocessor on your computer is responsible
  for the execution of your program.
• For any high-level language, when want to create
  an executable program, you use a compiler.
• Compiler ? Turns your high-level language
  statements into a format that the microprocessor
  can understand

10
Introduction Levels of Abstraction (2)

• Heres a nice drawing I made!
• Very simple example ? A B C at the high
  level
• On the microprocessor end, it does this addition
  by a form of instructions

11
Introduction Levels of Abstraction (3)

• Example of the use of instructions
• Cooking! Lets say you wanted to make scrambled
  eggs how would you do it?
• Crack eggs open and put them into bowl
• Whip and beat them using a fork
• Throw them on a pan and fry them

12
Introduction Levels of Abstraction (4)

• This is EXACTLY the same as what the
  microprocessor is trying to accomplish on its end
• Take a high level language piece of code and
  you break it down into instructions
• Processor carries out lines of code as a sequence
  of instructions

13
Introduction Levels of Abstraction (5)

• Lets look at the figure again
• You want to add the variables B and C and store
  them into the variable A
• You could do this by registers and memory
• All microprocessors that exist today have
  registers
• Registers ? Quick access memory storage elements
  for computation

14
Introduction Levels of Abstraction (6)

• So, look at the diagram again
• Procedure on how to do A B C
• Load a register with the value stored in memory
  location B
• Add the value of stored in memory location C to
  this register
• Store this new value to memory location A

15
Introduction Levels of Abstraction (7)

• But! The microprocessor can ONLY understand 0s
  and 1s! Look at the diagram again
• Once you compile, your program is turned into a
  series of instructions
• The assembler must now take these instructions
  and turn them into machine code (0s and 1s)

16
Introduction Levels of Abstraction (8)

• Look at the diagram again below
• Look at the instructions. You can see that there
  are two things associated to it The
  instruction, and the parameter for the
  instruction
• LDR B ? Use the load register command and load it
  with the contents of memory location B

17
Introduction Levels of Abstraction (9)

• An instruction requires two pieces of information
  in order to execute
• Opcode The actual instruction itself (LDR, STR)
• Operand The parameter to this instruction (A,
  B, C)
• This is pretty much like a function in C or Java
  ? int number gimmeSome(value)

18
Introduction Levels of Abstraction (10)

• You could have instructions that have no operands
  at all, just the opcode.
• C or Java equivalent spaceCadet()
• Now back to the figure
• Look at the machine code part of the diagram
• Takes the OpCode and Operand and turns them into
  0s and 1s
• Now, the microprocessor is ready to run the
  program

19
Introduction Levels of Abstraction (11)

• Let me mess up your minds a bit
• Instructions like load, store, add, etc these
  also need to perform instructions to perform the
  instructions! These are called
  micro-instructions
• Heres your time to go huh?????
• Back to cooking example
• Scrambled Eggs Youre performing
  micro-instructions without even knowing it!

20
Introduction Levels of Abstraction (12)

• First step ? Crack eggs and put them into a bowl.
  You need to do small things (microinstructions)
  in order to perform this step. What are they?
• Get eggs out of fridge
• Get bowl out of cupboard
• Place both on the table
• Hit eggs against surface
• Crack open and place eggs into bowl

21
Introduction Levels of Abstraction (13)

• Second step ? Whip and beat the eggs with a fork.
  Whatre the microinstructions?
• Get the fork out of the cupboard
• Grab a hold of the bowl
• Use the fork and whip the eggs
• Third step ? Dump eggs into frying pan
• Lift and hold the bowl
• Turn it upside down and dump onto the frying pan

22
Introduction Levels of Abstraction (14)

• So, what youre doing when youre cooking
  scrambled eggs is that youre following
  instructions, but performing micro-instructions
• Lets throw this all together
• High Level Language ? Open up a menu, tell the
  waiter you want scrambled eggs
• Assembly Language ? Chef makes scrambled eggs for
  you following instructions, performing
  micro-instructions
• Machine Code ? 00101 001011, 11010 100111, 11011
  100011, etc. etc.

23
Outline of Lecture

• Objectives
• Introduction
• Levels of abstraction
• Basic Computer Architecture
• Overview of Customized Ryerson Microprocessor
  (CRM)
• Datapath Unit on CRM
• Architecture
• Control Signals

24
Introduction Basic Computer Architecture

• The microprocessors in your computer, iPod, Cell
  Phone, PlayStationetc. have 2 very important
  elements that make it up
• Datapath Unit
• Handles manipulation and the movement of data in
  the microprocessor
• Responsible for all mathematical and logical
  operations performed on variables
• Has memory storage elements such as cache,
  registers, Static RAM

25
Introduction Basic Computer Architecture (2)

• Control Unit
• How exactly do you tell the datapath unit that
  you want to perform an add, store, load?
• Control Unit provides the proper signals that are
  fed to the datapath unit so it can perform the
  required operation
• Example Remote Control and TV
• Remote Control ? Control Unit
• TV performing the desired operation ? Done by
  Datapath Unit
• So! Control Unit tells the datapath on what it
  should do and the Datapath Unit does the actual
  grunt work itself

26
Outline of Lecture

• Objectives
• Introduction
• Levels of abstraction
• Basic Computer Architecture
• Overview of Customized Ryerson Microprocessor
  (CRM)
• Datapath Unit on CRM
• Architecture
• Control Signals

27
Overview of the CRM - Introduction

• Back in 1995 or so before
• I knew what the hell a computer was
• You all knew what the hell a computer was
• I was still sleeping at night with a night
  light. Oops.
• Ryerson profs developed a customized
  Microprocessor for this course so you guys know
• How a microprocessor works
• The basic architecture of a microprocessor
• How to program programs at the microprocessor
  level
• To make your lives miserable because they
  probably werent married and want to take it out
  on you guys

28
Overview of the CRM Introduction (2)

• They created and implemented their circuit
  designs for their customized microprocessor
• CRM is physically implemented
• Circuit is printed on a PCB (Printed Circuit
  Board) and is now ready for use
• Lets look at the datapath unit (dont piss your
  pants the circuit might look ugly)

29
Overview of the CRM Datapath Unit Architecture

• Heres what it looks like!
• Remember, Datapath does the grunt work
• What do you see here?
• Register (ACCA)
• ALU
• SRAM

30
Overview of the CRM Datapath Unit
Architecture (2)

• Big Vertical Solid Line ? Data Bus
• Mechanism used to transfer data between entities
• 4 bits wide
• Only 2 things can be connected to the data bus at
  one time
• Transmitter
• Receiver
• If you dont follow this youll get a data
  clashing!

31
Overview of the CRM Datapath Unit
Architecture (3)

• Register (Accumulator)?
• Call this accumulator because you have the
  ability to change the contents based on
  arithmetic or logical operations
• Basically stores numbers after computation
• Very special kind of register, a shift register.
• Can use to load in a number, and you can also
  shift bits around, either to the left or the right

32
Overview of the CRM Datapath Unit
Architecture (4)

• Arithmetic / Logic Unit?
• Used to do all arithmetic and logical operations
  on the microprocessor
• Takes in 2 inputs, does logical or arithmetic
  operations on them and stores them into the
  accumulator
• Output of the ALU is sent to the accumulator /
  shift register

33
Overview of the CRM Datapath Unit
Architecture (5)

• Tri-State Bus Driver?
• Acts basically like a switch
• Enables and Disables Output that its connected
  to
• 4 bits wide

34
Overview of the CRM Datapath Unit
Architecture (6)

• Input Switches?
• Can load numbers provided by you as well instead
  of just memory!
• Switches take in an input number from the user
  and goes to the ALU for computation purposes
• Connected to tri-state buffer to ensure no data
  clashing with other entities

35
Overview of the CRM Datapath Unit
Architecture (7)

• Static RAM?
• Used to store numbers from the switches or the
  accumulator
• Used to retrieve stored numbers for use in
  computing values
• Output of RAM also serves as input to the ALU
  (note the Chip Select used for no data clashing)
• 16 memory locations available for this Static RAM
  chip

36
Overview of the CRM Datapath Unit Control
Signals

• Tri-State Bus Driver Control Signal Inputs?
• For the outputs of the accumulator and getting
  data from the switches, there is a control switch
  used to enable or disable the outputs
• /AS Accumulator Select
• SM Switches or Memory
• /AS enables or disables the output of the
  accumulator
• SM disables or enables the switches

37
Overview of the CRM Datapath Unit Control
Signals (2)

• Static RAM Control Signal Inputs ?
• SM is used to disable and enable the RAM chip
• This signal is shared between the RAM and the
  Switches
• Note the active low / active high inversion on
  the switches and the RAM
• /WE enables writing to a memory location or
  reading from a memory location on the chip
• N3-0 selects the particular address in memory you
  want to access or write to.

38
Overview of the CRM Datapath Unit Control
Signals (3)

• Accumulator Control Signal Inputs ?
• Note the Q3 and Q0, the MSB and LSB of the ACCA
  output!
• C is for a carry-in into the register.
• Since youre performing arithmetic operations,
  you need a Carry-In, just like Lab 3
• CLK ? Very important signal. Youll see next
• S1 and S0 control the way the contents of the
  accumulator are manipulated
• S1 0, S0 0 ? Do not change the contents of
  the accumulator
• S1 0, S0 1 ? Shift bits to the right at each
  clock cycle
• S1 1, S0 0 ? Shift bits to the left at each
  clock cycle
• S1 1, S0 1 ? Do a direct loading from D3-0

39
Overview of the CRM Datapath Unit Control
Signals (4)

• ALU Control Signal Inputs
• A and B are inputs into the ALU. You perform
  Arithmetic and Logic Operations on these two
  inputs
• S3-0 are used to select the particular operation
  you want to perform on A and B
• /CALU Carry in into ALU (Same as Lab 3)
• M selects whether you want to perform Arithmetic
  (M 0) or Logic (M 1)
• /Co Carry out from ALU
• AEB - Flag goes high when B A 0

40
Outline of Lecture

• Objectives
• Introduction
• Levels of abstraction
• Basic Computer Architecture
• Overview of Customized Ryerson Microprocessor
  (CRM)
• Datapath Unit on CRM
• Architecture
• Control Signals

41
Outline of Lecture (2)

• Instruction set on CRM
• Details on each instruction available
• Control Unit on CRM
• Architecture
• Control Signals
• Procedure for Final Project
• Preparing EPROM Files
• Microinstructions ? EPROM Format
• Microinstruction Tables and Examples
• Format for code and addressing
• Conversion of EPROMs ? .asc to .hex

42
Overview of the CRM Instruction Set

• Instruction Set The set of all instructions
  that you can perform on a microprocessor
• The list of available instructions on the CRM is
  listed on page 5 of the lab7.pdf
• Each instruction in the CRM needs
• Opcode Actual instruction to perform
• (Optional) Operand Parameter to the instruction

43
Overview of the CRM Instruction Set (2)

• ADDA N Adds the accumulator with contents of
  memory location N and stores into accumulator
  (with the input carry to the accumulator)
• OpCode 0000
• Operand N3N2N1N0 ? 4 bit address in memory you
  want to access
• SUBA N Subtracts the accumulator with contents
  of memory location N and stores into accumulator
  (with the borrow bit to the accumulator)
• OpCode 0001
• Operand N3N2N1N0 ? 4 bit address in memory you
  want to access

44
Overview of the CRM Instruction Set (3)

• INPA N Takes a 4-bit binary number from the
  switches and puts this to the accumulator
• OpCode 0010
• Operand None (Dont Cares ? N3N2N1N0 xxxx)
• LDAA N Loads the accumulator with contents of
  memory location N
• OpCode 0011
• Operand N3N2N1N0 ? 4 bit address in memory you
  want to access

45
Overview of the CRM Instruction Set (4)

• STAA N Store the contents of the accumulator
  and store it into memory location N.
• OpCode 0100
• Operand N3N2N1N0 ? 4 bit address in memory you
  want to access
• JMP N Jump to instruction number (address) N in
  your program
• OpCode 0101
• Operand N3N2N1N0 ? 4 bit address in your program
  that you want to jump to
• Note, to implement a program using this
  instruction, you can only have maximum 16 lines
  of code!

46
Overview of the CRM Instruction Set (5)

• Something you need to know ? We could have two
  instructions with the same OpCode, but have the
  Operand be used to distinguish instructions from
  one another! (Dont cares are 1 (x 1))
• ADDA S Adds the accumulator with the input
  number seen at the switches and stores into
  accumulator (with the input carry to the
  accumulator)
• OpCode 0110
• Operand N3N2N1N0 ? xxx0 ? N0 0 to perform
  addition
• SUBA S Subtracts the accumulator with the input
  number seen at the switches and stores into
  accumulator (with the borrow bit to the
  accumulator)
• OpCode 0110
• Operand N3N2N1N0 ? xxx1 ? N0 1 to perform
  subtraction

47
Overview of the CRM Instruction Set (6)

• ANDA N Performs the logical AND operation with
  the accumulator and whatever contents are in
  memory location N and stores the result in the
  accumulator
• OpCode 0111
• Operand N3N2N1N0 ? 4 bit address in memory you
  want to access
• CLC Clears the input carry bit in the
  accumulator
• OpCode 1000
• Operand N3N2N1N0 ? xxx0 ? N0 0 to perform
  clear

48
Overview of the CRM Instruction Set (7)

• SEC Sets the input carry bit in the accumulator
• OpCode 1000
• Operand N3N2N1N0 ? xxx1 ? N0 1 to perform set
• DECA Decrement the accumulator (ACCA ACCA
  1)
• OpCode 1001
• Operand N3N2N1N0 ? xxx0 ? N0 0 to perform
  decrement

49
Overview of the CRM Instruction Set (8)

• INCA Increment the accumulator (ACCA ACCA
  1)
• OpCode 1001
• Operand N3N2N1N0 ? xxx1 ? N0 1 to perform
  increment
• RORA Rotate the Accumulator 1 bit to the right
  (MSB of accumulator C and LSB of accumulator
  C)
• OpCode 1010
• Operand N3N2N1N0 ? xxx0 ? N0 0 to perform
  right rotate

50
Overview of the CRM Instruction Set (9)

• ROLA Rotate the Accumulator 1 bit to the left
  (LSB of accumulator C and C MSB of
  accumulator)
• OpCode 1010
• Operand N3N2N1N0 ? xxx1 ? N0 1 to perform left
  rotate
• STSW N ? Take a number from the input switches
  and store it into memory location N
• This combines two operations INPA and STAA N
• OpCode 1011
• Operand
• 1st step ? Operand xxxx ? Number to place into
  accumulator comes from switches
• 2nd step ? N3N2N1N0 ? Memory location to store
  input switch value to

51
Overview of the CRM Instruction Set (10)

• Jump Instructions ? When you write an assembly
  program, youre able to jump to any instruction
  in your program
• Doesnt have to follow the exact sequence line by
  line
• JCS N Jump to instruction number N when the
  carry bit is set
• OpCode 1100
• Operand N3N2N1N0 ? Address (instruction number)
  to jump to if carry is set
• JEQ N Jump to instruction number N when the
  accumulator (A) equals the switches (B)
• OpCode 1101
• Operand N3N2N1N0 ? Address (instruction number)
  to jump to A B

52
Overview of the CRM Instruction Set (11)

• JMI N Jump to instruction number N when the
  number is negative (MSB of accumulator 1)
• OpCode 1110
• Operand N3N2N1N0 ? Address (instruction number)
  to jump to MSB of accumulator 1
• CLRA Clear the contents of the accumulator
• OpCode 1111
• Operand N3N2N1N0 ? xxx0 ? N0 0 to perform
  clear
• SETA Set all accumulator bits to 1
• OpCode 1111
• Operand N3N2N1N0 ? xxx1 ? N0 1 to perform set

53
Outline of Lecture (2)

• Instruction set on CRM
• Details on each instruction available
• Control Unit on CRM
• Architecture
• Control Signals
• Procedure for Final Project
• Preparing EPROM Files
• Microinstructions ? EPROM Format
• Microinstruction Tables and Examples
• Format for code and addressing
• Conversion of EPROMs ? .asc to .hex

54
Overview of the CRM Prequel - EPROM

• So, what the hells an EPROM?
• EPROM ? Electronically Programmable Read Only
  Memory
• Similar to RAM ? Write stuff to memory locations
• Can also access values that you wrote to this
  chip by an address

55
Overview of the CRM Prequel EPROM (2)

• However, once you write, you cannot overwrite
  like you do for PALs!
• When you write stuff to it and you make a
  mistake, you must erase all of the contents all
  at once!
• Silver square in the middle is susceptible to UV
  light.
• You must place the EPROM under a UV light eraser
  for about 10 15 minutes to clear the contents

56
Overview of the CRM Control Unit Architecture

• Heres what it looks like! If you didnt piss
  your pants, nows the time
• Remember, Control Unit tells the Datapath on how
  it should perform instructions
• What do you see here?
• EPROMs
• PALs
• Program Counter
• Register

57
Overview of the CRM Control Unit Architecture
(2)

• Program Counter ? Keeps track of the next
  instruction number you will be running in your
  code
• The SW7 switch (CLR), resets your Program Counter
  to 0 so that you start at the beginning of your
  program

58
Overview of the CRM Control Unit Architecture
(3)

• EPROMs 1 and 2 ? For a particular instruction you
  want to perform, these store the microinstruction
  signals required to perform the instruction you
  want.
• How this works
• Opcode serves as input to EPROM (there are more
  inputs, but well look at those later)
• Microinstruction sequence number also serves as
  input (need to know which microinstruction you
  want to execute)
• These are used to find the memory location of the
  proper control signals needed
• Output is the actual control signals themselves

59
Overview of the CRM Control Unit Architecture
(4)

• Microinstruction Next State Register
• This keeps track of the next microinstruction you
  need to perform for a given instruction
• Remember, an instruction requires
  microinstructions in order to execute properly
• Once you complete a microinstruction, this tells
  you what microinstruction you need to perform
  next
• Fed back to EPROMs 1 and 2 to figure out next
  microinstruction

60
Overview of the CRM Control Unit Architecture
(5)

• EPROM 3 Stores programs in assembly format for
  the microprocessor to execute
• 2 ways to create programs
• C or Java way
• Assembly way (This is the way were going to make
  programs)
• EPROM 3 stores the instructions for up to 4
  programs in this EPROM
• Each program must be max 16 lines long

61
Overview of the CRM Control Unit Architecture
(6)

• PAL
• Used to output additional control signals for a
  microinstruction for a given instruction
• EPROMs do not have the ability to give output
  based on combinatorial logic.
• These provide unconditional output You give a
  memory location, you get an output
• There are certain instructions that require
  control signals to be generated based on certain
  conditions that need to be met
• PAL generates conditional control signals (Jump
  Instructions!)

62
Overview of the CRM Recap of Computer
Architecture

• Remember, Control Unit provides control signals
  that are fed to the datapath.
• Datapath uses these control signals to perform
  what you want it to do!
• Do you guys want a break?
• YES ? 10 minute break
• NO ? Well keep going
• Which one???

63
Overview of the CRM Control Unit Control
Signals

• Program Counter Control Signal Generation
• Receives CLR from you when you want to reset the
  program counter
• CLK is also received from you. You tell the
  microprocessor you want to go to the next
  instruction by this signal (well get into this
  later)
• PA3-0 provide the next instruction to be executed
  in the program (connected to LEDs!)
• N3-0 provide the jump address if you want to jump
  to a particular instruction number

64
Overview of the CRM Control Unit Control
Signals (2)

• Program Counter Control Signal Generation (2)
• /PE and CNT are very special
• /PE 1 and CNT 0 Execute the next
  microinstruction for the instruction
• /PE 1 and CNT 1 Increment the program
  counter and go to the next instruction
• /PE 0 and CNT 0 or 1 You want to jump to an
  address in your program given by N3-0
• Remember, N3-0 is also the operand part of the
  instruction
• When you want to jump, you specify the address of
  where you want to jump here

65
Overview of the CRM Control Unit Control
Signals (3)

• EPROM 1 and 2 Control Signal Generation
• Inputs
• OpCode (O3-O0) Used to figure out what
  instruction you want to execute
• A1-0 Used to figure out what microinstruction
  you want to execute for the given OpCode
• N0 Used to figure out what instruction you want
  to execute if two instructions share the same
  OpCode
• A7 Used to tell these EPROMs if a jump condition
  has been met (more on this later)

66
Overview of the CRM Control Unit Control
Signals (4)

• EPROM 1 and 2 Control Signal Generation
• EPROM 1 is used to generate control signals for
  the ALU per microinstruction
• EPROM 1 Outputs
• S3-0 Outputs the function you want the ALU to
  execute
• M Arithmetic or Logic Operation select
• S1-0 Shift register control select
• NCC If you dont expect the output carry to
  change, set this to 1, else 0.

67
Overview of the CRM Control Unit Control
Signals (5)

• EPROM 1 and 2 Control Signal Generation
• EPROM 2 is used to generate control signals for
  memory access, data bus control and Program
  Counter control
• Also provides the next microinstruction number to
  execute
• EPROM 2 Outputs
• D1-0 Provides the next microinstruction to
  execute
• CNT Used to increment Program Counter
• /PE Used to toggle when you want to jump to an
  address or not
• SM Used to switch between memory or switches
• WR Used to read from or write to memory
• /AS Used to disable and enable accumulator
  output

68
Overview of the CRM Control Unit Control
Signals (6)

• EPROM 3 Signal Generation
• Inputs
• You need PA3-0 to figure out the next instruction
  in your program you want to execute
• SW4 and SW5 These are input switches that tell
  the microprocessor what program you want to run
• SW5 0, SW4 0 ? 1st program
• SW5 0, SW4 1 ? 2nd program
• SW5 1, SW4 0 ? 3rd program
• SW5 1, SW4 1 ? 4th program
• Note Offsetting!

69
Overview of the CRM Control Unit Control
Signals (7)

• EPROM 3 Signal Generation
• Outputs
• O3-0 output the OpCode of the particular
  instruction to be executed, given the address
  from the Program Counter
• N3-0 output the Operand of the instruction
• Note This also is fed to the program counter
  because Jump Instructions use the Operand to
  specify a jump address
• Also, the N0 bit is fed to EPROM1 and 2 for
  instructions with the same OpCode
• Theres a slight problem with this N0 bit
  feeding. Youll see later

70
Overview of the CRM Control Unit Control
Signals (8)

• Microinstruction Next State Register Control
  Signal Generation
• Inputs and Outputs
• CLR is used to reset the register so that the
  next microinstruction to run will be the first.
• CLK is provided by you so it will tell EPROMs 1
  and 2 what the next microinstruction is at a
  clock cycle
• Every clock cycle is 1 microinstruction
• D1-0 is used to grab the next microinstruction
  number from EPROM 2 and store it
• AEB is the A equals B flag provided by the ALU
  given the state of the previous version AEBD
• C is the next carry provided to the accumulator
  based on the previous carry C
• A1-0 outputs whatever was given to D1-0 at the
  next clock cycle

71
Overview of the CRM Control Unit Control
Signals (9)

• PAL Control Signal Generation
• Inputs
• Q0 LSB of ACCA
• Q3 MSB of ACCA
• O3-0 OpCode given by EPROM 3
• CLK Provided by you as earlier
• N0 As described earlier
• WR Used to enable reading or writing for memory
• NCC Set to 0 if you want to change the input
  carry to ACCA, else 1
• /Co Carry out from the ALU goes here to
  determine conditional control signals
• AEBD Comes from Next-State Register
• C Like above

72
Overview of the CRM Control Unit Control
Signals (10)

• PAL Control Signal Generation
• Outputs
• /WE This asserts a write to memory only when WR
  1, AND when the CLK is low to ensure that you
  can write properly (dont worry about this too
  much)
• C - Determines the next state of the carry
• /CALU Determines the carry-in into the ALU
• A7 VERY special
• A7 is 1 ONLY when a jump condition has been
  satisfied
• When A B for JEQ N
• When carry is set for JCS N
• When MSB of ACCA is 1 for JMI N
• JMP N jump signal is not generated here because
  it is not a conditional jump

73
Procedure for Final Project More about the PAL

• Take a look at page 6 of Lab 7
• You can see that for the PAL, theyve already
  generated the PAL equations for you!
• You must
• Look at the pinouts for the PAL on the Control
  Unit diagram and create a PALASM file based on
  these inputs
• Remember PIN 10 GND, PIN 20 VCC, and any
  pin thats not described in the diagram NC (No
  connect)
• Write out the logic equations you see for each
  output pin in PALASM format

74
Outline of Lecture (2)

• Instruction set on CRM
• Details on each instruction available
• Control Unit on CRM
• Architecture
• Control Signals
• Procedure for Final Project
• Preparing EPROM Files
• Microinstructions ? EPROM Format
• Microinstruction Tables and Examples
• Format for code and addressing
• Conversion of EPROMs ? .asc to .hex

75
Procedure for Final Project What the CRM looks
like

• 4 places where you plug in your 3 EPROMs and 1
  PAL
• Datapath and Control Unit are already physically
  implemented
• Switch Interface is used to control the
  microprocessor behaviour
• SW3-0 are the input switches
• PA5-4 are the program selection bits
• CLR is used to reset the program counter to 0
• Remember to toggle on and off!

76
Procedure for Final Project So now what?

• So, whats already given to you
• Instruction Set with OpCode and Operand formats
• Datapath Unit physically implemented
• Control Unit physically implemented
• So, whatre you supposed to do now?
• What they DIDNT give you is the behaviour of the
  microinstructions
• They dont tell you what control signals should
  be asserted or cleared for a given instruction
• Its YOUR job to figure out which signals are
  high or low for every microinstruction per
  instruction!

77
Procedure for Final Project Microinstructions

• Remember, each instruction can take up to 4
  microinstructions to implement it
• However, most of the instructions given in this
  instruction set can be implemented in JUST 1
  microinstruction
• The only instructions that require 2
  microinstructions are
• JEQ N
• STSW N

78
Procedure for Final Project Microinstructions
(2)

• For each microinstruction, there is a total of 15
  control signals generated by EPROMs 1 and 2 to be
  sent to the datapath unit for processing
• Microinstructions will assert the proper control
  signals needed for the datapath unit to properly
  execute the given instruction at hand
• These microinstructions will be stored in EPROMs
  1 and 2

79
Procedure for Final Project Microinstructions
(3)

• How does microinstruction signal generation work?
• We need to use
• OpCode
• Microinstruction in the sequence you want to run
• The jump indication bit A7
• Shared OpCode indication bit N0
• These need to be used to access EPROMs 1 and 2 in
  order output the proper control signals for the
  right microinstruction to execute properly

80
Procedure for Final Project Microinstructions
(4)

• Lets create something new
• Lets combine all of these inputs into a single
  8-bit input address to EPROMs 1 and 2 which will
  be formatted as follows
• A7A6A5A4A3A2A1A0
• A7 Jump Indication Bit
• A6 N0 ? Shared OpCode Indication Bit
• A5 A4 A3 A2 OpCode of the instruction you want
• A1A0 Microinstruction number of the instruction
  you want executed
• Example If A7A6A5A4A3A2A1A0 00000000, this
  means that
• A7 0 ? Conditional Jump Condition has not been
  met
• A6 0 ? Shared OpCode Indiciation Bit 0 ?
  Choosing the first instruction of the two
  instructions which share the same OpCode
• A5A4A3A2 0000 ? We want to perform the ADDA N
  operation
• A1A0 00 ? From the ADDA N operation, we want to
  perform the first microinstruction needed for
  this instruction

81
Procedure for Final Project Microinstructions
(5)

• Given this new method for access microinstruction
  controls in the EPROMs, there are 4 possibilities
  that you guys will encounter
• A7 0, A6 (N0) 0 When the jump condition
  hasnt been met and you want to possibly execute
  the first instruction of the shared OpCode set
  (CLC, SUBA S, RORA, DECA, CLRA)
• A7 0, A6 (N0) 1 When the jump condition
  hasnt been met and you want to possibly execute
  the second instruction of the shared OpCode set
  (SEC, ADDA S, ROLA, INCA, SETA)
• A7 1, A6 (N0) 0 When the jump condition HAS
  been met and you want to possibly execute the
  first instruction of the shared OpCode set
• A7 1, A6 (N0) 1 When the jump condition HAS
  been met and you want to possibly execute the
  second insruction of the shared OpCode set

82
Procedure for Final Project Microinstructions
(6)

• Why possibly? Remember what I said when I was
  looking at the control unit and said N0 would
  create a problem?
• Because of us making N0 part of the combined
  8-bit addressing vector, lets say we want to
  execute an instruction like ADDA N, or LDAA N and
  so on
• Remember, N0 is used for shared OpCode
  instructions and this is the LSB of the Operand
• But, if we want to use ADDA N or LDAA N, N0 could
  be anything!

83
Procedure for Final Project Microinstructions
(7)

• N0 when used by these kinds of instructions
  states the memory location that you want to use,
  so N0 can equal 0 or 1
• What does this mean? Lets do an example
• If we know that
• The jump condition isnt satisfied (A7 0)
• The instruction we want to perform is ADDA N
• The microinstruction we want to execute is the
  first one (A1A0 00)
• This means we could have
• A7A6A5A4A3A2A1A0 00000000 if we wanted to
  access memory location 0 (N0 0) for example.
• But what if we wanted to do the above, but
  access memory location 1?
• A7A6A5A4A3A2A1A0 01000000 if we wanted to
  access memory location 1 (N0 1)

84
Procedure for Final Project Microinstructions
(8)

• N0 should not affect the control signals to the
  datapath required to perform
• Access to memory
• Disabling and Enabling the switches or memory
• Accumulator Disabling and Enabling
• So, regardless of N0, we want to access THE SAME
  microinstruction control signals
• If N0 0 or 1, we should still generate the same
  control signal values!
• What do we do here? At A7A6A5A4A3A2A1A0
  00000000 and at A7A6A5A4A3A2A1A0 01000000, we
  should have duplicate entries in the EPROM
• In both of these memory locations in the EPROM,
  we should output the SAME microinstruction
  control signals!

85
Procedure for Final Project Microinstructions
(9)

• So, instructions that do not use the operand to
  distinguish between instructions need to have
  their microinstruction signals duplicated for 2
  different addresses
• Something to note ? Jump Condition Bit, A7, will
  NEVER go high when calling instructions like ADDA
  N, SUBA N etc.
• Jump condition bit only goes high when you call
  the conditional jump instructions and the
  condition to jump is true
• For the instructions that DO use the operand to
  distinguish between instructions, their
  microinstruction control signals will be
  obviously different

86
Procedure for Final Project Microinstructions
(10)

• Example Lets do the CLRA and SETA instructions
• CLRA ? Opcode 1111, Operand xxx0 1110SETA ?
  Opcode 1111, Operand xxx1 1111
• Assume were at the first microinstruction for
  both of these instructions
• So, for CLRA ? A7A6A5A4A3A2A1A0 00111100 ? Use
  this memory location to access control signals
  for CLRA
• For SETA ? A7A6A5A4A3A2A1A0 01111100 ? Use this
  memory location to access control signals for
  SETA
• These different memory locations SHOULD have
  different microinstruction control signals
  because they are DIFFERENT instructions!

87
Procedure for Final Project Microinstructions
(11)

• Next, suppose we want to execute a jump
  instruction
• Lots of different cases thatll happen
• Example 1 JMP N
• LSB of ACCA is used to specify part of jump
  address, so N0 can be both 0 and 1.
• A7 can never be 1 for this. Only 1 when
  conditional jump condition is satisfied
• Opcode 0101 Operand Any 4-bit number
• Assume first microinstruction
• So, A7A6A5A4A3A2A1A0 00010100 will be the
  address to access the microinstruction signals
  for both EPROMs 1 and 2 for N0 0
• Should be the same microinstruction signals for
  when N0 1 as well

88
Procedure for Final Project Microinstructions
(12)

• Example 2 JCS N
• LSB of ACCA is used to specify part of jump
  address, so N0 can be both 0 and 1.
• A7 will be either 0 when a condition to jump
  hasnt been met, and 1 when a condition to jump
  has been met
• Opcode 1100 Operand Any 4-bit number
• Assume first microinstruction
• Case 1 A7 0
• In this case, A7 will equal 0, therefore
  A7A6A5A4A3A2A1A0 000110000 for N0 0. The
  microinstruction control signals should simply
  move to the next instruction ? Jump Condition
  hasnt been met.
• Same for when N0 1

89
Procedure for Final Project Microinstructions
(13)

• Case 2 A7 1
• In this case, A7 will equal 1, therefore
  A7A6A5A4A3A2A1A0 000110000 for N0 0. The
  microinstruction control signals should be
  asserted so that the program counter should load
  the instruction address to jump to
• Same for when N0 1 the same logic as last time

90
Procedure for Final Project Microinstructions
(14)

• Now that youre aware of the things we need to
  know about microinstruction control signal
  generation, its time to figure out when to
  assert these signals for each microinstruction
• Take a look on Pages 17 19
• They give you nicely laid out tables for you to
  fill out the proper microinstruction signals to
  assert for a given instruction
• Remember, there are 15 different control signals
  coming out from both EPROMs 1 and 2

91
Procedure for Final Project Microinstruction
Tables

• Heres what part of the table looks like
• 3 different columns exist for each EPROM
• 1st column Shows your 8-bit combined input
  addressing vector shows all combinations
  possible with this combined addressing
• Note the conversion of this combined address to
  hexadecimal
• 2nd column Shows you the microinstruction
  control signals asserted for the 8-bit combined
  input addressing for this row for EPROM 1
• 3rd column Shows you the microinstruction
  control signals asserted for the 8-bit combined
  input addressing for this row for EPROM 2

92
Procedure for Final Project Microinstruction
Tables (2)

• EPROM 1 ? Control signals that are outputted from
  here given the 8-bit combined input address are
• NCC No change to next input carry of ACCA
• S1 and S0 Control the shift register
• M ALU Arithmetic and Logic Operation Control
• S3-0 Control what specific ALU operation you
  want

93
Procedure for Final Project Microinstruction
Tables (3)

• EPROM 2 ? Control signals that are outputted from
  here given the 8-bit combined input address are
• D7 Dont know what this is and I cant
  remember. Always set this to dont care (1)
• /AS Accumulator Disable / Enable
• WR Reads from memory or writes to memory
• /PE Program Counter uses or ignores jump
  address fed in
• CNT Used to determine whether to execute a
  microinstruction or an instruction
• A1, A0 - Used to specify the next
  microinstruction to execute

94
Procedure for Final Project Microinstruction
Tables (4)

• You will notice that there are 4 tables that you
  need to fill out
• There are 4 tables because of the 4 scenarios
  involving A7 and N0
• Hints
• Control signals for the instructions that dont
  use N0 for shared OpCode instructions will have
  the same signals for when N0 0 and N0 1
• Those that do use N0 for shared OpCode
  instructions WILL be different for when N0 0
  and N0 1
• Control signals for instructions that attempt a
  conditional jump and fail should only make the
  microprocessor go to the next instruction (A7
  0)

95
Procedure for Final Project Microinstruction
Tables (5)

• When A7 1, this means that a conditional jump
  condition is true
• Less entries for this table since A7 can never be
  1 for instructions like ANDA N, SUBA S, etc.
• When A7 1, you should assert the proper control
  signals for this microinstruction which will load
  the program counter with whatever is in the
  Operand N3-0
• When A7 1, the tables should be the same for
  both N0 0 and 1.
• N0 is used as part of the address to jump to.
  Duplicate microinstruction signals are required

96
Procedure for Final Project Examples

• Lets do ADDA N (Page 13)
• EPROM 1 Signals
• NCC 0 since carry to the accumulator can change
  (arithmetic operation youll see!)
• S1 S0 1 since you want to load the register
  with some contents
• M 0 for arithmetic operation
• S3-0 1001 ? Specifies adding 2 numbers (Page 4
  of sn74ls181.pdf)
• EPROM 2 Signals
• D7 Dont Care (1)
• /AS 1 since you want to disconnect the
  accumulator with the data bus
• WR 0 since you want to read from memory address
  N
• SM 0 since you want to choose memory and not
  the switches
• /PE 1 since you want to ignore jumping to an
  address
• CNT 1 since you want the program counter to
  increment next instruction to execute
• A1 0 and A0 0 to point to the next
  microinstruction to execute (00 for the next
  instruction)

97
Procedure for Final Project Examples (2)

• Lets do JEQ N (Page 13)
• First, we must test to see if A B or B A 0
• EPROM 1 Signals
• NCC 1 The carry of the accumulator shouldnt
  change. Not performing any arithmetic operations
  to store into accumulator. Only testing
• S1 0, S0 0 The accumulator should not
  change its contents were just testing if B A
  0
• M 0 for arithmetic instruction
• S3-0 0110 specifies subtracting ? A B
• EPROM 2 Signals
• D7 1 as before
• /AS 1 since we are testing for B A. Should
  not be connected to data bus
• WR 0 to be safe. Were not using memory, but
  safer to read than write to avoid memory glitches
• SM 1 to choose switches
• /PE 1 to ignore jump address
• CNT 0 to execute next microinstruction
  specified by next-state register
• A1 0, A0 1, to execute the second
  microinstruction

98
Procedure for Final Project Examples (3)

• The next microinstruction to execute will depend
  on when the A equals B flag is 0 or 1
• Case 1 When A equals B flag 0
• Therefore A7 0, and we should just go to the
  next instruction
• EPROM 1 Signals
• NCC 1 so carry isnt changed again
• S1 S0 0 since we dont want to change ACCA
  contents
• M and S3-0 are dont cares (1) since we just want
  to go to next instruction
• EPROM 2 Signals
• D7 Dont Care (1)
• /AS 1 to disable attachment to data bus for
  safety
• WR 0 since its safer to read than write for
  this case
• SM Dont care (1) because it doesnt matter
  whether you want to enable the switches or
  memory you just want the next instruction
• /PE 1 to ignore jump address
• CNT 1 since you want to execute next
  instruction
• A1 A0 0 so you can execute first
  microinstruction of next instruction

99
Procedure for Final Project Examples (4)

• Case 2 When A equals B flag 1
• Therefore A7 1, which means we need to load the
  program counter with the address of N and jump
  there
• EPROM 1 Signals
• NCC 1 so carry isnt changed again
• S1 S0 0 since we dont want to change ACCA
  contents
• M and S3-0 are dont cares (1) since we just want
  to go to next instruction specified by N
• EPROM 2 Signals
• D7 Dont Care (1)
• /AS 1 to disable attachment to data bus for
  safety
• WR 0 since its safer to read than write for
  this case
• SM Dont care (1) because it doesnt matter
  whether you want to enable the switches or
  memory you just want the next instruction
• /PE 0 to use jump address
• CNT Dont care because it doesnt matter
  whether you want to go to next instruction or
  microinstruction. Instruction number WILL change
• A1 A0 0 so you can execute first
  microinstruction of next instruction

100
Procedure for Final Project Examples (5)

• Final example STSW N
• Does 2 instructions INPA and STAA N
• Step 1 INPA
• EPROM 1 Signals
• NCC 0 since carry to the accumulator can change
  (arithmetic operation youll see!)
• S1 S0 1 since you want to load the register
  with some contents
• M 1 for logical operation
• S3-0 1010 ? The output of the ALU will be
  simply B, the switches (more below)
• EPROM 2 Signals
• D7 Dont Care (1)
• /AS 1 since you want to disconnect the
  accumulator with the data bus
• WR 0 since you want to be safe and set it to
  read. Writing might cause glitches
• SM 0 since you want to choose the switches and
  not memory
• /PE 1 since you want to ignore jumping to an
  address
• CNT 1 since you want to execute the next
  microinstruction specified by below
• A1 0 and A0 1 to point to the second
  microinstruction to execute

101
Procedure for Final Project Examples (6)

• Step 2 STAA N
• EPROM 1 Signals
• NCC 1 since carry to the accumulator shouldnt
  change. Only storing.
• S1 S0 0 since contents of register shouldnt
  change
• M and S3-0 Dont Cares ? Not using ALU. Were
  only storing
• EPROM 2 Signals
• D7 Dont Care (1)
• /AS 0 since you want to connect the accumulator
  with the data bus to transfer to memory
• WR 1 since you want to write from memory
  address N
• SM 0 since you want to choose memory and not
  the switches
• /PE 1 since you want to ignore jumping to an
  address
• CNT 1 since you want the program counter to
  increment next instruction to execute
• A1 0 and A0 0 to point to the next
  microinstruction to execute (00 for the next
  instruction)